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Chapter 4: Problem 6

Let \(A\) and \(B\) be arbitrary sets. Complete the following statements. (a) \(A \subseteq B \Leftrightarrow A \cap B=\) (b) \(A \subseteq B \Leftrightarrow A \cup B=\) (c) \(A \subseteq B \Leftrightarrow A-B=\) (d) \(A \subset B \Leftrightarrow(A-B=\) \(-\) ). (e) \(A \subset B \Leftrightarrow(A \cap B=\) \(\wedge A \cap B \neq\) (f) \(A-B=B-A \Leftrightarrow\)

Short Answer

Expert verified
(a) \(A\); (b) \(B\); (c) \(\emptyset\); (d) \(\emptyset\); (e) \(A \neq B\); (f) \(A = B\).

Step by step solution

01

Understanding Subset Definition

A set \(A\) is a subset of a set \(B\), denoted \(A \subseteq B\), if every element of \(A\) is also an element of \(B\). To solve this, we express set operations (intersection, union, difference) in terms of subset relations.
02

Statement (a): Intersection Equals A

The intersection \(A \cap B\) represents the elements that both \(A\) and \(B\) have in common. If \(A \subseteq B\), then \(A \cap B\) must equal \(A\) because all elements of \(A\) are shared with \(B\). Thus, \(A \subseteq B \Leftrightarrow A \cap B = A\).
03

Statement (b): Union Equals B

The union \(A \cup B\) represents all elements that are in \(A\), in \(B\), or in both. If \(A \subseteq B\), all elements of \(A\) are in \(B\), so the union must be \(B\). Thus, \(A \subseteq B \Leftrightarrow A \cup B = B\).
04

Statement (c): Set Difference is Empty

The set difference \(A - B\) includes all elements of \(A\) that are not in \(B\). If \(A\) is a subset of \(B\), \(A - B\) should be empty, denoted by \(\emptyset\). So, \(A \subseteq B \Leftrightarrow A - B = \emptyset\).
05

Statement (d): Set Difference is Non-Empty Proper Subset

A subset \(A\) of \(B\) is a proper subset, denoted \(A \subset B\), if \(A\) is a subset of \(B\) and not equal to \(B\). This implies \(A - B = A\) but \(A\) is strictly smaller than \(B\), so \(\emptyset eq A - B = A\).
06

Statement (e): Intersection and Non-Empty Difference

For \(A \subset B\), \(A \cap B = A\) but, since \(A\) is a proper subset, there has to be something exclusive to \(B\), ensuring \(A \cap B eq B\). Thus, \(A \subset B \Leftrightarrow (A \cap B = A \wedge A \cap B eq B)\).
07

Statement (f): Complement Symmetry

The expression \(A - B = B - A\) is evaluated for its possibilities. This is true only when \(A = B\) because the elements not mutually shared are isolated in both expressions. Thus, \(A - B = B - A \Leftrightarrow A = B\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Subsets
A subset is one of the fundamental concepts in set theory. A set \( A \) is considered a subset of set \( B \), symbolized as \( A \subseteq B \), if every element of \( A \) is also an element of \( B \). The idea here is simple:
  • All members of \( A \) are found in \( B \).
  • If this holds true, then we say \( A \) is contained within \( B \).
If there's at least one element in \( A \) that is not in \( B \), then \( A \) is not a subset of \( B \). In practical terms, you can imagine an everyday item being inside a box - if the item fits entirely, then the item is a subset of the box's contents.
Set Operations
Set operations allow us to explore relationships and interactions between different sets. There are three primary operations you will often encounter:
  • Intersection (\( A \cap B \)): This operation finds the common elements shared by sets \( A \) and \( B \). If \( A \subseteq B \), then every element in \( A \) must also be in \( B \), and hence \( A \cap B = A \).
  • Union (\( A \cup B \)): This operation combines all elements from both \( A \) and \( B \). If \( A \subseteq B \), the union results in \( B \), as all elements of \( A \) are already included in \( B \).
  • Difference (\( A - B \)): This operation focuses on elements unique to \( A \) after removing common elements with \( B \). If \( A \subseteq B \), then \( A - B \) will be empty, \( \emptyset \), because no different elements exist only in \( A \).
Proper Subsets
A proper subset is a more exclusive classification. A set \( A \) is a proper subset of a set \( B \), denoted as \( A \subset B \), if:
  • \( A \) is a subset of \( B \) \((A \subseteq B)\).
  • \( A \) and \( B \) are not equal, meaning \( B \) has at least one element not found in \( A \).
For example, consider \( A = \{1, 2\} \) and \( B = \{1, 2, 3\} \). Here, every element of \( A \) is in \( B \), yet \( B \) has an additional element \( 3 \). Thus, \( A \) is a proper subset of \( B \).
Set Difference
The set difference \( A - B \) represents elements that are unique to set \( A \), after excluding any elements that are also present in set \( B \). It's like checking what remains in \( A \) if all shared elements with \( B \) are removed.
  • If \( A \subseteq B \), then \( A - B = \emptyset \), as all elements of \( A \) are contained within \( B \).
  • For a proper subset \( A \subset B \), the condition \( A - B = A \) cannot be true, since \( A eq \emptyset \) and \( A \) is not equal to \( B \).
  • In symmetrical set differences, \( A - B = B - A \), occurs only when \( A = B \). Any difference implies a distinction in membership.
Understanding the set difference helps clarify how sets relate regarding exclusive membership.

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Most popular questions from this chapter

Give examples of sets \(A\), \(B\) and \(C\) such that: (a) \(A \subset B\) and \(B \subset C,\) and \(A \notin C\) (b) \(A \in B\) and \(B \in C,\) and \(A \in C\)

Let \(A\) and \(B\) be arbitrary nonempty sets. (a) Under what condition does \(A \times B=B \times A ?\) (b) Under what condition is \((A \times B) \cap(B \times A)\) empty?

Let \(A, B,\) and \(C\) be any three sets. Prove that (a) \(A \times(B \cap C)=(A \times B) \cap(A \times C)\) (b) \(A \times(B-C)=(A \times B)-(A \times C)\)

Comment on the following statements. Are they syntactically correct? (a) \(x \in A \cap x \in B \equiv x \in A \cap B\) (b) \(x \in A \wedge B \Rightarrow x \in A \cap B\)

Let \(A, B,\) and \(C\) be any three sets. Prove that if \(A \subseteq B\), then \(A \times C \subseteq B \times C\).
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